Basic analysis: proof for commutativity of exponents I am self-studying Rudin's textbook on mathematical analysis (3rd edition). In question $6$ of chapter $1$:

Fix $b>1$. If $m$, $n$, $p$, $q$ are integers, $n>0$, $q>0$, and $r=m/n=p/q$, prove that
$$(b^m)^{1/n} = (b^p)^{1/q}.$$
Hence it makes sense to define $b^r = (b^m)^{1/n}$.

A solution I read defines a unique real number $c$ such that $c^{nq} = b^k$, then computes $((b^m)^{1/n})^{nq} = b^k$, and the same for $((b^p)^{1/q})^{nq} = b^k$. Thus the statement is proved.
My problem is, since this proof involves basic computation of exponents (i.e. $(b^m)^{1/n} = b^{m/n})$, why instead of simply computing on $(b^m)^{1/n}$ and $(b^p)^{1/q}$, it was necessary to define/point out a unique $c$ and take the 'long route'?
edit: Thank you all for the comments and answers so far! It seems that including the proof I read may clarify my confusion, so here it is.

Let $k = mq = np$. Since there is only one positive real number $c$ such that $c$$nq$$ = b$$k$ (By a theorem), if we prove both ($b$$m$)$1/n$ and ($b$$p$)$1/q$ have this property, it will follow that they are equal. The proof is then a routine computation: (($b$$m$)$1/n$)$nq$ = ($b$$m$)$^q = b^{mq} = b^k$, and similarly for $(b^p)^{1/q}$.

Isn't this 'routine computation' what is taught in high school mathematics? How is this proof different from a mechanical execution such rules?
 A: The premise of the problem is that we have no idea what $ b^{m/n} $ even is, and we're setting out to come up with a reasonable definition for it.  So when you point out that the inference $ (b^m)^{1/n} = b^{m/n} $ is a "basic computation of exponents," you're missing the point.  You are setting out to show that it makes sense to define $ b^{m/n} = (b^m)^{1/n} $.  In order to show that it "makes sense", you need to verify the equation $ (b^m)^{1/n} = (b^p)^{1/q} $ without making any use whatsoever of the formula $ b^{m/n} = (b^m)^{1/n} $.
The proof you cited succeeds in this; at no point does it make direct reference to any number called $ b^{m/n} $.  You actually skipped some steps in their computation (maybe because you didn't fully understand it):
$$
((b^m)^{1/n})^{nq} = (((b^m)^{1/n})^n)^q = (b^m)^q = b^{mq}
$$
The critical step is the third '$ = $' sign, where we use the fact that $ (R^{1/n})^n = R $, with $ R = b^m $.
A: I'm not sure I understand your question or maybe I don't understand the answer being quoted.
We know that $b^m$ ($m$ and integer and $b>1$) exists and is $b^m >1$.  And the definition of $W^{\frac 1n}$ (for  $W>1$) is the unique positive real $V$ so that $V^n= W$ (and we know that a unique such real exists by a theorem) so
$(b^m)^{\frac 1n}$ is equal to the unique positive number $d$ so that $d^n = b^m$.  We must define such a $d$ because .... we have no other way of defining $(b^m)^{\frac 1n}$.
Likewise $(b^p)^{\frac 1q}$ is the unique positive number $\delta$ so that $\delta^{q} = b^p$.
So that is your suggestion of finding $(b^m)^{\frac 1n}$ "directly".
But we have a task of proving that $d = \delta$. We can't avoid that task.  Ans we don't have any idea how to compare the values of $d$ and $\delta$ other than knowing that $d^n=b^m$ and $\delta^q =b^p$.
And so $d^{nq}=(d^{n})^q = b^{mq}$ and $(\delta^{q})^n = \delta^{nq}=b^{pn}$.  But $mq = np$ so $d^{nq}=b^{mq}=b^{pn}=\delta^{nq}$ so $d^{nq}=\delta^{nq}$.
And by the theorem that for every value $b^{mq}$ then is a unique number $[b^{mq}]^{\frac 1{nq}}$ and that number must be $\delta$ and that number must be $d$.
So all I can tell is you cited answer decided to refer to the $[b^{mq}]^{\frac 1{nq}}$  by the variable $c$ and refer to the number $mq = np$ as $k$.
But the proof didn't need those extra variables.  For that matter, I didn't need $d$ or $\delta$.  I couldve done the prove in a single line as:

$[(b^m)^{\frac 1n}]^{nq} = b^{mq} = b^{np} = [(b^p)^{\frac 1q}]^{nq}$ so both $(b^m)^{\frac 1n}= (b^p)^{\frac 1q}= [b^{np}]^{\frac 1{nq}}$.

But that would have been too terse and abstract and as these are new concepts we are trying to convince ourselves what the mean it is best to express these concepts by what they mean rather than shuffling symbols on paper.
Maybe the answer cited tossed in two many symbols.  But both it and your suggestion both do the same thing so for as I can tell; you must calculate that both $(b^m)^{\frac 1n}$ and $(b^p)^{\frac 1q}$ are the same $nq$th root of the $b^{mq}=b^{np}$.
====addition======

Isn't this 'routine computation' what is taught in high school mathematics?

Maybe.  My experience in high school we took things for granted though.
For natural $n$ we were told $b^n$ is $b$ multiplied by itself $n$ times.  For natural numbers we could prove/demonstrate that $b^{n+m} =b^nb^m$ and $(b^n)^m=b^{nm}$ (that's just "counting"; more formally it's a result of the induction principle we didn't know that).
And we extended that definition it makes sense to define $b^w$ for negative numbers.  And this is fine as we know division works and is the inverse of multiplication.
But our first blind flying leap at the moon, was when we said: Since $\frac 1n \times n = 1$ and $(b^n)^m = b^{nm}$ we should define $b^{\frac 1n}$ as $\sqrt[n]{b}$ so that way we'd have $(b^{\frac 1n})^n = b^{\frac 1n\cdot n}=b^1=b$.
Why is that blind flying leap at the moon?  Because we don't know that there is any such number that raised to the $n$th power will be $b$ or that it is unique. (in fact if $n$ is even then there is none if $b$ is negative, and if $b$ is positive and $n$ is even there are two such numbers.
And then our huge leap is assuming therefore we we can define $b^{\frac mn}$ so $\sqrt[n]{b^m}$.  Why is that a leap?  Consider $0.4 = \frac 4{10} = \frac 25$.  Do we know that $\sqrt[10]{b^4} = \sqrt[5]{b^2}$?  Do we know that those numbers even exist?  And we do we assume that all ore rules about $b^{r+s}=b^rb^s$ are going to still work?
So that is why we are doing this exercise.
Theorem 1.21 says that for $b>1$ and natural $n$ there is a unique positive $c$ so that $c^n=b$ and we proved that.  We called that number $b^{\frac 1n}$ with obvious foresight that that $(b^{\frac 1n})^n = b^{\frac 1n\times n}$ but we don't know that just because the notation looks the same and we have one case where our rules work we have no reason to think this is consistent.
Now we know that $(b^{\frac 1n})$ exists and is a positive number and we know that $(b^{\frac 1n})^m$ exists and is a postive number.  So we could try to define, for any two natural numbers $(n,m)$ then $b^{\frac nm}:= (b^{\frac 1n})^m$.  The only problem is $\frac nm\leftrightarrow (n,m)$ is not one to one.  $\frac nm$ is a single rational value and if $\frac nm = \frac pq$ that does not mean that $(n,m) = (p,q)$ and to define $b^{\frac nm}$ as $(b^{\frac 1m})^n$ we must have that if $\frac mn=\frac pq$ we must have $b^{\frac mn} =b^{\frac pq}$ which means we must have $(b^{\frac 1m})^n = (b^{\frac 1m})^n$.
So that is our task. If we can prove that for any pair of numbers $(n,m)$ and $(p,q)$ so that $\frac nm = \frac pq$ that  the, arguably arbitrary, manipulations of $(b^{\frac 1m})^n$ and $(b^{\frac 1q})^p$ both yield the same result then we will be allowed to define for $r \in \mathbb Q$ so that $r = \frac ac; a\in \mathbb Z; c\in \mathbb N$ then $b^r := (b^{\frac 1c})^a$.
When I was in high school we took all of that for granted and didn't worry about confirming out assumptions.
